Question

Let A be the area of a triangle with semi- perimeter p. Show that A cannot be greater than p^3/3√3.

Answer 1

Question: Let A be the area of a triangle with semi- perimeter p. Show that A cannot be greater than p^3/3√3.

$A \leq p^3/3\sqrt3$

Proceeding from the fact that we are not given which triangle it is. And the ratio of the area of the triangle to its perimeter is simply given. Therefore, to determine the area and perimeter, an equilateral triangle can be used to facilitate the calculation. Because it doesn't matter which sides of the triangle.

Consider an equilateral triangle with sides $а(a=b=c)$ , then:

Area of a triangle: $A= \frac{a*b*sinC}{2}\Rightarrow A = \frac{a^2*sin60}{2} =a^2*\frac{\sqrt3}{4}$

Half-perimeter equals: $p=\frac{a+b+c}{2} \Rightarrow p=\frac{3a}{2} \Rightarrow a=2p/3$

Substitute the values of $а$ from the half-perimeter formula into the triangle area formula.

$A= (\frac{2p}{3})^2*\frac{\sqrt3}{4}=\frac{4p^2}{9}*\frac{\sqrt3}{4}=\frac{p^2\sqrt3}{9}\Rightarrow$

$\frac{p^2\sqrt3}{9}=\frac{p^2\sqrt3}{\sqrt81}=\frac{p^2}{3\sqrt3}$

As you can see $A=\frac{p^2}{3\sqrt3}$ , and this: $\frac{p^2}{3\sqrt3}\leq \frac{p^3}{3\sqrt3}$

This inequality will work for any triangles. Any area of a triangle will be less than $\frac{p^3}{3\sqrt3}$

From this it follows that A cannot be greater than $\frac{p^3}{3\sqrt3}$ .